What are the solutions to the equation -16t^2-32t+1395=0 ?

The solutions to the equation -16t^2-32t+1395=0 are t=-(4+sqrt(1411))/4 ,t=(sqrt(1411)-4)/4

Find the zeros of -16t^2-32t+1395=0

The zeros of -16t^2-32t+1395=0 are t=-(4+sqrt(1411))/4 ,t=(sqrt(1411)-4)/4

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- 16 t^{2} - 32 t + 1395 = 0

t = - \frac{4 + 1411}{4}, t = \frac{1411 - 4}{4}

Decimal Notation

t = - 10.39081 \dots, t = 8.39081 \dots

Solution steps

- 16 t^{2} - 32 t + 1395 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 ( - 16 ) \cdot 1395}{2 ( - 16 )}

(- 32)^{2} - 4 (- 16) \cdot 1395 = 81411

t_{1, 2} = \frac{- ( - 32 ) \pm 8 1411}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- ( - 32 ) + 8 1411}{2 ( - 16 )}, t_{2} = \frac{- ( - 32 ) - 8 1411}{2 ( - 16 )}

t = \frac{- ( - 32 ) + 8 1411}{2 ( - 16 )} : - \frac{4 + 1411}{4}

t = \frac{- ( - 32 ) - 8 1411}{2 ( - 16 )} : \frac{1411 - 4}{4}

The solutions to the quadratic equation are:

t = - \frac{4 + 1411}{4}, t = \frac{1411 - 4}{4}

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What are the solutions to the equation -16t^2-32t+1395=0 ?
The solutions to the equation -16t^2-32t+1395=0 are t=-(4+sqrt(1411))/4 ,t=(sqrt(1411)-4)/4
Find the zeros of -16t^2-32t+1395=0
The zeros of -16t^2-32t+1395=0 are t=-(4+sqrt(1411))/4 ,t=(sqrt(1411)-4)/4